Lowest Common Ancestor of Words

题目来源
这是一个LCA的问题 ，要自己建树比较麻烦。
既然是自己建树，那就自己想怎么建怎么建了，我只建parent pointer就好了，根本不需要child pointer。
用HashMap表示。
找的时候可以先把一个到根的路径找出来放到hashSet里面，再走另一个，第一个遇到的重复的节点就是了。
也可以看谁离根最远，先走最远的那一条路径，然后两个路径一起走，遇到相同的就返回。这样就省了空间。
但是建树的空间还是不能省的 。

public class LCAWords {

    Map<String, String> parentMap;
    public LCAWords(String[][] words) {
        parentMap = new HashMap<>();
        for (String[] record : words) {
            String p = record[0];
            for (int i = 1; i < record.length; i++) parentMap.put(record[i], p);
        }
    }

    public String find(String w1, String w2) {
        Set<String> w1Parent = new HashSet<>();
        String w1n = w1;
        while (w1n != null) {
            w1Parent.add(w1n);
            w1n = parentMap.get(w1n);
        }
        while (w2 != null) {
            if (w1Parent.contains(w2)) return w2;
            w2 = parentMap.get(w2);
        }
        return null;
    }
    public String find2(String w1, String w2) {
        int n1 = distToParent(w1, parentMap);
        int n2 = distToParent(w2, parentMap);
        while (n1 > n2) {
            w1 = parentMap.get(w1);
            n1--;
        }
        while (n2 > n1) {
            w2 = parentMap.get(w2);
            n2--;
        }
        while (!w1.equals(w2)) {
            w1 = parentMap.get(w1);
            w2 = parentMap.get(w2);
        }
        return w1;
    }
    private int distToParent(String w, Map<String, String> parentMap) {
        int n = 0;
        while (w != null) {
            w = parentMap.get(w);
            n ++;
        }
        return n;
    }
    public static void main(String[] args) {
        String[][] words = new String[][] {{"earth", "north america", "south america"},
                {"south america", "brazil", "arginta"},
        {"north america", "united states", "canada"},
            {"united states", "california", "new york"},
                {"california", "san francisco", "oakland"}};
        String word1 = "california";
        String word2 = "oakland";
        LCAWords solultion = new LCAWords(words);
        Display.myPrint(solultion.find(word1, word2));
        Display.myPrint(solultion.find2(word1, word2));

    }
}